You are at a game show and there are three closed doors. There is a prize hidden behind one of the doors and the game show host knows where it is. You are asked to choose a door. The game show host then opens one of the other two doors showing that it is empty and asks you if you would like to change your selection. Should you stick to your original selection?

There is a barrel with no lid and some beer in it. "This barrel is more than half full," said Chuck. "No it's not," say Joe. "It's less than half full." Without any measuring implements and without removing any beer from the barrel, how can they easily determine who is correct?

I fail to see what will change whether you switch or not. Imagine being asked which door to choose and being shown whats behind the one door before choosing. You will have %50 chance of being correct on either choice. Any other answer will be concluded out of "fake" logic.

Although having thought about it a little, if you understand human psychology a little, you would be able to read the game show host and determine where it is from his choice. Although this still has nothing to do with logic.

I'm not sure what you are saying, and i'm not sure if it is undebatable with this definition of the question, but the principle is that when he shows you an empty door, he cuts one of the 1/3 doors away resulting in a 2/3 door and a 1/3 door left. The funny part is that so many mathematics raged against this answer, until finally in the end giving up realising they were wrong.

Yes i didn't mention.. He choses one of the doors you did not already select.. You still hold on to your 1/3 door while he choses one of the other two 1/3, transfering the 1/3 to the door that is left :)

How do you transfer odds? You're making this complicated while it's not.
3 doors = every door is 1/3. Remove one door and choose again: 2 doors = every door is 1/2. What am I missing here? :>
Choosing again does not imply changing your selection (but might). You see 2 doors and you choose - staying with 'your' door _is_ choosing in my eyes.

ahh yes I remember that back from maths in school.
You have a 2/3 chance of picking an empty door first try while the host will definitely show you one of them afterwards. So by switching and not staying you do have 2/3 odds in your favor because you only lose if you picked the prize before.

I never really liked it because it implies some things to work.
- the host would have to know where the prize is.
- the host does offer the chance of switching every time not only if you picked the prize firsthand.
- the stuff behind the doors would've to be 'fixed'. When watching these shows as a kid I always had the feeling they were switching stuff around anyway because of 1) :>

As I said "fake" logic. I have read this before, however, maths is not correct in everything. I'm sure you've seen the series of equations proving 0 = 1? "fake" logic.

What I am trying to say is that whether you have chosen the correct door or not the host will still show you an empty door, which means it is a choice between 2 doors all along, because you will always know of one empty door.

punee's example clarified pretty well how the odds for having made a wrong decision on the first guess are high, because there were more possible wrong alternatives. the situation with two doors is 50:50, but the probability that the door you chose in the first place is the wrong one is 2/3.

and "proofs" like 0 = 1, a = b or whatever are simply wrong math. they include illegal mathematical operations at one point or another all of the time.

math is always correct in, or (in the case of stochastics) provides the most probable solution for everything.

It is more serious thinking than math, that's why so many math people claimed it wrong. I still don't know quite what you are saying, so let me ask you this: Are you able to see the situation where switching door raises the probability from 1/3 to 2/3?

(notify: I am very drunk at this moment) At this moment, I am not working forward, I am working backward.Imagine being given a choice between 2 doors... that is exactly that you are being given in this problem.

OK, i will try to explain once more, follow this visualy in your head:

Imagine you pick one door: After you have picked a door, the host asks you if you want to pick the two-other-doors instead; so that, either you stick with your door, or you take the two-other-doors. As there is an equal probability that the prize is behind every door, and since the host always asks if you want to select the two-other-doors, the two-other-doors are surely better selection than the one you have chosen.

NOW, imagine that he opens one of the-two-other doors that is empty. As he always opens a door that is empty, the probability is still the same! Nothing changes! He just shows you the one that is empty, there is always one that is empty, he just shows you which one.

Got a lovely one for you, although if you read alot of puzzles you've probably heard it.

A doctor wants to test how intelligent his friends are so he invites 3 of them over and tells them "I have 5 hats, 3 white ones and 2 black ones, I am going to put a hat on each of your heads and put you in a room, you can only come out of the room once you are sure which colour hat you are wearing. You are not allowed to communicate in anyway"

So he blind folds them and leads them into the room and puts a white hat on each of their heads. He then takes off the blind folds and leaves the room.

One hour later one of his friends comes out the room and says "I a wearing a white hat" and takes off the hat. How did he come to this solution?

I've just read puzzle, can I take part in this one ? :-D (I love this kind of logical)
Instead of giving straight the answer (a quite long cascaded one), a few tips :

- it took a long of time before one of the friend comes out and telling the answer... maybe he was "waiting" for his friends behavior ?
- the one giving the answer must believe his two mates aren't smartards (if they were he could have found a wrong answer)

If you still don't get it, just think the "If I was wearing a black hat, what would think my mates about their own hat ?"

I have seen a similar puzzle but it was with 3 sentenced to death penalty sat down on a chair... with a last chance to survive if answering the right answer..... SAW anyone ? :->

Yeah i was thinking of that.. eg. if your mates are wearing two black hats, you would know you had white; or similiar if there were two black hats in the group, someone would know their own hat. Also, if there are one black and two white hats in the group, there will be a black and a white hat in front of someone, and he might asume that he does not have a black hat if noone leaves right away. But then again your white-hat mate might leave when you leave, causing trouble -- this is where i don't get it. If there are three white hats, noone could leave or asume anything besides that everyone has a white hat after a long time; because, if you had a black hat, one of the other two would say that they had a white one before long.

i could probably have crammed that into two lines, but that's how my thinking went :P

- A sees two white hats
- Either A has a black hat or he has a white hat
- If A has a black hat, then B and C see a white and a black hat.
- If B and C see a white and a black hat, and neither B nor C moves, then both B and C have a white hat (for if B moved, it would mean C has a black hat because B could infer that seeing two black hats, his must be white and vice versa)
- Nobody moves
- Therefore A's hat must be white.